What is the integral from 0 to 1 of 1/(1-x^4) ?

The integral from 0 to 1 of 1/(1-x^4) is diverges

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integral from 0 to 1 of 1/(1-x^4)

\int_{0}^{1} \frac{1}{1 - x ^{4}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{1 - x ^{4}} d x

Take the partial fraction of

\frac{1}{1 - x ^{4}} : \frac{1}{2 ( x ^{2} + 1 )} + \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x - 1 )}

= \int_{0}^{1} \frac{1}{2 ( x ^{2} + 1 )} + \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x - 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{1}{2 ( x ^{2} + 1 )} d x + \int_{0}^{1} \frac{1}{4 ( x + 1 )} d x - \int_{0}^{1} \frac{1}{4 ( x - 1 )} d x

\int_{0}^{1} \frac{1}{2 ( x ^{2} + 1 )} d x = \frac{π}{8}

\int_{0}^{1} \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln (2)

\int_{0}^{1} \frac{1}{4 ( x - 1 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 1 of 1/(1-x^4) ?
The integral from 0 to 1 of 1/(1-x^4) is diverges